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The tangent to the curve `y=x^2-5x+5.` parallel to the line `2y=4x+1,` also passes through the point :A. `((1)/(4), (7)/(2))`B. `((7)/(2), (1)/(4))`C. `(-(1)/(8), 7)`D. `((1)/(8), -7)` |
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Answer» Correct Answer - D The given curve is ` y = x ^(2) - 5x + 5 " " ` … (i) Now, slope of tangent at any point `(x, y)` on the curve is `" " (dy)/(dx) = 2x - 5" "`… (ii) `" " `[ on differentiating Eq. (i) w.r.t. x] `because` It is given that tangent is parallel to line ` " " 2y = 4x + 1 ` So, ` (dy)/(dx) = 2 [ because` slope of line ` 2y = 4x + 1 ` is 2`]` `rArr 2x - 5 = 2 rArr 2x = 7 rArr x = (7)/(2)` On putting `x= (7)/(2)` in Eq. (i), we get `" " y = (49)/( 4) - ( 35)/( 2) + 5 = ( 69)/( 4) -( 35)/( 2) = - (1)/(4)` Now, equation of tanent to the curve (i) at point `((7)/(2), - (1)/(4))` and having slope 2 , is `" " y + (1)/(4) = 2(x -(7)/(2)) rArr y + (1)/(4) = 2x - 7` `rArr " " y = 2x - (29)/(4)" "` ...(iii) On checking all the options, we get the point `((1)/(8), -7)` satisfy the line (iii). |
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